If F(1)=2 and F(n)=F(n-1)+1/2 for all integers n>1,then F(101)=?

Question

Please answer with detailed explanation

Answer 1

We can try and derive a closed form for the recursive series.
F(1) = 2
F(2) = F(1) + (1/2) = 2 + (1/2)
F(3) = F(2) + (1/2) = 2 + (1/2) + (1/2)
F(4) = F(3) + (1/2) = 2 + (1/2) + (1/2) + (1/2)

It appears the pattern is that there is (n - 1) of (1/2) added on to 2 for the n-th term. Therefore, the closed form expression seems to be

F(n) = 2 + (n - 1)(1/2)

Which means

F(101) = 2 + (101 - 1)(1/2)
F(101) = 2 + (100/2)
F(101) = 2 + 50 = 52

Detailed explanation proving by induction:
If F(1) = 2 and F(n) = F(n - 1) + (1/2), then
F(n) = 2 + (n - 1)(1/2)

1) Let n = 1. Then by definition, F(1) = 2, and
F(1) = 2 + (1 - 1)(1/2) = 2 + 0 = 2, so this works.

2) Assume the formula holds true for some value n = k. That is,

If F(k) = F(k - 1) + (1/2), then F(k) = 2 + (k - 1)(1/2).

Let's consider F(k + 1).
By definition,
F(k + 1) = F(k) + (1/2)

By our induction hypothesis, we know what F(k) is equal to.
In { } brackets, I will put F(k).

F(k + 1) = {2 + (k - 1)(1/2)} + (1/2)
F(k + 1) = 2 + (1/2)k - (1/2) + (1/2)
F(k + 1) = 2 + (1/2)k
F(k + 1) = 2 + (1/2)({k + 1} - 1)

This shows that the formula holds true for n = k + 1.
Since we proved that if the formula holds for n = k, that the formula holds true for n = k + 1, it follows, by the principle of mathematical induction, that

If F(1) = 2 and F(n) = F(n - 1) + (1/2), then
F(n) = 2 + (n - 1)(1/2)

for all natural numbers n.

Answer 2

Since 1/2 is added each time then it's just an arithmetic sequence. The formula for the nth term of an A.S. is well known.

Answer 3

52